Chapter 14: Probability
Exercise 14.1
Q1
Complete the following statements:
(i) Probability of an event E + Probability of the event 'not E' = ______.
(ii) The probability of an event that cannot happen is ______. Such an event is called ______.
(iii) The probability of an event that is certain to happen is ______. Such an event is called ______.
(iv) The sum of the probabilities of all the elementary events of an experiment is ______.
(v) The probability of an event is greater than or equal to ______ and less than or equal to ______.
(i) Probability of an event E + Probability of the event 'not E' = ______.
(ii) The probability of an event that cannot happen is ______. Such an event is called ______.
(iii) The probability of an event that is certain to happen is ______. Such an event is called ______.
(iv) The sum of the probabilities of all the elementary events of an experiment is ______.
(v) The probability of an event is greater than or equal to ______ and less than or equal to ______.
Answer:
(i) 1
(ii) 0, Impossible event
(iii) 1, Sure event (or Certain event)
(iv) 1
(v) 0, 1
Q2
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
(i) Not equally likely. The car starting depends on various factors (mechanical condition, fuel, etc.). It is not purely random like a coin toss.
(ii) Not equally likely. The outcome depends on the player's skill. A good player is more likely to score than miss.
(iii) Equally likely. There are only two possible answers (True or False), and without knowledge, each has an equal chance of being selected.
(iv) Equally likely. There are two possibilities (Boy or Girl), and biologically, they are considered equally likely outcomes.
Q3
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Solution:
Tossing a coin is considered fair because the possible outcomes are only two: Head or Tail. Assuming the coin is unbiased, these outcomes are equally likely. The result of an individual toss is completely unpredictable.
Q4
Which of the following cannot be the probability of an event?
(A) \( \frac{2}{3} \)
(B) -1.5
(C) 15%
(D) 0.7
(A) \( \frac{2}{3} \)
(B) -1.5
(C) 15%
(D) 0.7
Answer: (B)
Reason: The probability of an event always lies between 0 and 1 (inclusive). Probability cannot be negative. Therefore, -1.5 cannot be a probability.
Q5
If \( P(E) = 0.05 \), what is the probability of 'not E'?
Solution:
We know that \( P(E) + P(\text{not } E) = 1 \).
\[ P(\text{not } E) = 1 - P(E) \]
\[ P(\text{not } E) = 1 - 0.05 = 0.95 \]
Q6
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out:
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution:
(i) The bag contains only lemon flavoured candies. There are no orange flavoured candies.
This is an impossible event.
\[ P(\text{orange flavoured candy}) = 0 \]
(ii) Since all candies are lemon flavoured, taking out a lemon candy is a certain event.
\[ P(\text{lemon flavoured candy}) = 1 \]
Q7
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let \( E \) be the event that 2 students have the same birthday.
Then, \( \text{not } E \) is the event that they do not have the same birthday.
Given \( P(\text{not } E) = 0.992 \).
We know that \( P(E) + P(\text{not } E) = 1 \).
\[ P(E) = 1 - P(\text{not } E) \]
\[ P(E) = 1 - 0.992 = 0.008 \]
Q8
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?
Solution:
Number of red balls = 3
Number of black balls = 5
Total number of balls = \( 3 + 5 = 8 \)
(i) Probability of drawing a red ball:
\[ P(\text{Red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{3}{8} \]
(ii) Probability of drawing a ball that is not red:
\[ P(\text{Not Red}) = 1 - P(\text{Red}) = 1 - \frac{3}{8} = \frac{5}{8} \]
Q9
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?
Solution:
Total marbles = \( 5 \text{ (Red)} + 8 \text{ (White)} + 4 \text{ (Green)} = 17 \).
(i) Probability of Red:
\[ P(\text{Red}) = \frac{5}{17} \]
(ii) Probability of White:
\[ P(\text{White}) = \frac{8}{17} \]
(iii) Probability of Not Green:
Outcomes favorable to 'not green' are Red and White marbles: \( 5 + 8 = 13 \).
\[ P(\text{Not Green}) = \frac{13}{17} \]
Q10
A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin? (ii) will not be a ₹ 5 coin?
Solution:
Number of 50p coins = 100
Number of ₹ 1 coins = 50
Number of ₹ 2 coins = 20
Number of ₹ 5 coins = 10
Total coins = \( 100 + 50 + 20 + 10 = 180 \).
(i) Probability of a 50p coin:
\[ P(50p) = \frac{100}{180} = \frac{5}{9} \]
(ii) Probability of not a ₹ 5 coin:
Number of coins that are not ₹ 5 = \( 100 + 50 + 20 = 170 \).
\[ P(\text{not ₹ 5}) = \frac{170}{180} = \frac{17}{18} \]
Q11
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 14.4). What is the probability that the fish taken out is a male fish?
Solution:
Number of male fish = 5
Number of female fish = 8
Total number of fish = \( 5 + 8 = 13 \)
Probability of selecting a male fish:
\[ P(\text{Male}) = \frac{5}{13} \]
Q12
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 14.5), and these are equally likely outcomes. What is the probability that it will point at:
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
Solution:
Total outcomes = 8 (Numbers 1, 2, 3, 4, 5, 6, 7, 8)
(i) Point at 8:
Favorable outcome: {8} (1 outcome)
\[ P(8) = \frac{1}{8} \]
(ii) An odd number:
Favorable outcomes: {1, 3, 5, 7} (4 outcomes)
\[ P(\text{Odd}) = \frac{4}{8} = \frac{1}{2} \]
(iii) A number greater than 2:
Favorable outcomes: {3, 4, 5, 6, 7, 8} (6 outcomes)
\[ P(>2) = \frac{6}{8} = \frac{3}{4} \]
(iv) A number less than 9:
All numbers 1 to 8 are less than 9. (8 outcomes)
\[ P(<9) = \frac{8}{8} = 1 \]
Q13
A die is thrown once. Find the probability of getting:
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
Solution:
Total outcomes on a die = {1, 2, 3, 4, 5, 6} (6 outcomes)
(i) A prime number:
Prime numbers on a die: {2, 3, 5} (3 outcomes)
\[ P(\text{Prime}) = \frac{3}{6} = \frac{1}{2} \]
(ii) A number lying between 2 and 6:
Numbers: {3, 4, 5} (3 outcomes)
\[ P(\text{Between 2 and 6}) = \frac{3}{6} = \frac{1}{2} \]
(iii) An odd number:
Odd numbers: {1, 3, 5} (3 outcomes)
\[ P(\text{Odd}) = \frac{3}{6} = \frac{1}{2} \]
Q14
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds
Solution:
Total cards = 52.
(i) A king of red colour:
Red kings are King of Hearts and King of Diamonds (2 cards).
\[ P(\text{Red King}) = \frac{2}{52} = \frac{1}{26} \]
(ii) A face card:
Face cards are Jack, Queen, King for all 4 suits. \( 3 \times 4 = 12 \) cards.
\[ P(\text{Face Card}) = \frac{12}{52} = \frac{3}{13} \]
(iii) A red face card:
Face cards in Hearts and Diamonds. \( 3 + 3 = 6 \) cards.
\[ P(\text{Red Face Card}) = \frac{6}{52} = \frac{3}{26} \]
(iv) The jack of hearts:
There is only 1 Jack of Hearts.
\[ P(\text{Jack of Hearts}) = \frac{1}{52} \]
(v) A spade:
There are 13 Spades in a deck.
\[ P(\text{Spade}) = \frac{13}{52} = \frac{1}{4} \]
(vi) The queen of diamonds:
There is only 1 Queen of Diamonds.
\[ P(\text{Queen of Diamonds}) = \frac{1}{52} \]
Q15
Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
Total cards = 5 (Ten, Jack, Queen, King, Ace).
(i) Probability of Queen:
There is 1 Queen.
\[ P(\text{Queen}) = \frac{1}{5} \]
(ii) Queen put aside:
Total remaining cards = 4 (Ten, Jack, King, Ace).
(a) An ace:
There is 1 Ace in the remaining 4 cards.
\[ P(\text{Ace}) = \frac{1}{4} \]
(b) A queen:
Since the Queen was put aside, there are 0 Queens left.
\[ P(\text{Queen}) = \frac{0}{4} = 0 \]
Q16
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Number of defective pens = 12
Number of good pens = 132
Total pens = \( 12 + 132 = 144 \)
Probability of getting a good pen:
\[ P(\text{Good}) = \frac{\text{Number of good pens}}{\text{Total pens}} = \frac{132}{144} \]
\[ = \frac{11}{12} \]
Q17
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
(i) Total bulbs = 20. Defective = 4.
\[ P(\text{Defective}) = \frac{4}{20} = \frac{1}{5} \]
(ii) The first bulb drawn was not defective (it was good), and it is not replaced.
Total remaining bulbs = \( 20 - 1 = 19 \).
Total defective bulbs remains = 4.
Total good bulbs remaining = \( 16 - 1 = 15 \).
Probability that the new bulb is not defective (good):
\[ P(\text{Not Defective}) = \frac{15}{19} \]
Q18
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number (ii) a perfect square number (iii) a number divisible by 5.
Solution:
Total discs = 90.
(i) A two-digit number:
Two-digit numbers are from 10 to 90. Count = \( 90 - 9 = 81 \).
\[ P(\text{Two-digit}) = \frac{81}{90} = \frac{9}{10} \]
(ii) A perfect square number:
Squares: \( 1, 4, 9, 16, 25, 36, 49, 64, 81 \). (9 outcomes).
\[ P(\text{Perfect Square}) = \frac{9}{90} = \frac{1}{10} \]
(iii) A number divisible by 5:
Multiples of 5: \( 5, 10, \dots, 90 \). Count = \( \frac{90}{5} = 18 \).
\[ P(\text{Divisible by 5}) = \frac{18}{90} = \frac{1}{5} \]
Q19
A child has a die whose six faces show the letters as given below:
A, B, C, D, E, A
The die is thrown once. What is the probability of getting (i) A? (ii) D?
A, B, C, D, E, A
The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:
Total outcomes = 6.
(i) Probability of getting A:
'A' appears on 2 faces.
\[ P(A) = \frac{2}{6} = \frac{1}{3} \]
(ii) Probability of getting D:
'D' appears on 1 face.
\[ P(D) = \frac{1}{6} \]
Q20
Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter 1m?
Solution:
Length of rectangle \( l = 3 \) m, Breadth \( b = 2 \) m.
Area of rectangle = \( 3 \times 2 = 6 \text{ m}^2 \).
Diameter of circle = 1 m, so Radius \( r = 0.5 \) m.
Area of circle = \( \pi r^2 = \pi (0.5)^2 = 0.25\pi \text{ m}^2 \).
Probability = \( \frac{\text{Area of circle}}{\text{Area of rectangle}} \)
\[ P = \frac{0.25\pi}{6} = \frac{\pi}{24} \]
Q21
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it?
(ii) She will not buy it?
(i) She will buy it?
(ii) She will not buy it?
Solution:
Total pens = 144.
Defective pens = 20.
Good pens = \( 144 - 20 = 124 \).
(i) She will buy it (Pen is good):
\[ P(\text{Buy}) = \frac{124}{144} = \frac{31}{36} \]
(ii) She will not buy it (Pen is defective):
\[ P(\text{Not Buy}) = \frac{20}{144} = \frac{5}{36} \]
Q22
Refer to Example 13. (i) Complete the following table:
(Sum on 2 dice vs Probability)
(ii) A student argues that 'there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11'. Do you agree with this argument? Justify your answer.
(Sum on 2 dice vs Probability)
(ii) A student argues that 'there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11'. Do you agree with this argument? Justify your answer.
Solution:
(i) Completed Table:
| Sum | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Probability | \(\frac{1}{36}\) | \(\frac{2}{36}\) | \(\frac{3}{36}\) | \(\frac{4}{36}\) | \(\frac{5}{36}\) | \(\frac{6}{36}\) | \(\frac{5}{36}\) | \(\frac{4}{36}\) | \(\frac{3}{36}\) | \(\frac{2}{36}\) | \(\frac{1}{36}\) |
(ii) No, the argument is incorrect.
Justification: The 11 sums are not equally likely. For example, sum '2' can only occur in 1 way (1,1), whereas sum '7' can occur in 6 ways (1,6; 2,5; 3,4; 4,3; 5,2; 6,1).
Q23
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Possible outcomes for 3 tosses: HHH, HHT, HTH, THH, TTH, THT, HTT, TTT.
Total outcomes = 8.
Winning outcomes (same result): {HHH, TTT} (2 outcomes).
Losing outcomes: \( 8 - 2 = 6 \).
\[ P(\text{Losing}) = \frac{6}{8} = \frac{3}{4} \]
Q24
A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
Solution:
Total outcomes = \( 6 \times 6 = 36 \).
Outcomes where 5 comes up at least once:
(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)
(5,1), (5,2), (5,3), (5,4), (5,6)
Total favorable outcomes for (ii) = 11.
(ii) Probability 5 comes up at least once:
\[ P(E) = \frac{11}{36} \]
(i) Probability 5 will not come up either time:
This is the complement of (ii).
\[ P(\text{Not 5}) = 1 - \frac{11}{36} = \frac{25}{36} \]
Q25
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3.
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.
(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3.
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.
Solution:
(i) Incorrect.
Reason: When two coins are tossed, the possible outcomes are {HH, HT, TH, TT}. These are 4 equally likely outcomes. 'One of each' includes {HT, TH}, which has a probability of \( \frac{2}{4} = \frac{1}{2} \), not \( \frac{1}{3} \).
(ii) Correct.
Reason: The outcomes on a die are {1, 2, 3, 4, 5, 6}. Odd numbers are {1, 3, 5} (3 outcomes) and even numbers are {2, 4, 6} (3 outcomes). Since both events have equal chances (3 out of 6), the probability is indeed \( \frac{1}{2} \).